Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $n = \dfrac{x^2 - 7x}{x^2 - 8x + 7} \times \dfrac{2x - 2}{x - 5} $
First factor the quadratic. $n = \dfrac{x^2 - 7x}{(x - 1)(x - 7)} \times \dfrac{2x - 2}{x - 5} $ Then factor out any other terms. $n = \dfrac{x(x - 7)}{(x - 1)(x - 7)} \times \dfrac{2(x - 1)}{x - 5} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ x(x - 7) \times 2(x - 1) } { (x - 1)(x - 7) \times (x - 5) } $ $n = \dfrac{ 2x(x - 7)(x - 1)}{ (x - 1)(x - 7)(x - 5)} $ Notice that $(x - 7)$ and $(x - 1)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ 2x(x - 7)\cancel{(x - 1)}}{ \cancel{(x - 1)}(x - 7)(x - 5)} $ We are dividing by $x - 1$ , so $x - 1 \neq 0$ Therefore, $x \neq 1$ $n = \dfrac{ 2x\cancel{(x - 7)}\cancel{(x - 1)}}{ \cancel{(x - 1)}\cancel{(x - 7)}(x - 5)} $ We are dividing by $x - 7$ , so $x - 7 \neq 0$ Therefore, $x \neq 7$ $n = \dfrac{2x}{x - 5} ; \space x \neq 1 ; \space x \neq 7 $